1/x+1/x+2≤1/x+1

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- Nov 5th 2008, 04:44 PMhogersolve the inequality
1/x+1/x+2≤1/x+1

- Nov 5th 2008, 05:10 PMPlato
Solve this inequally

- Nov 7th 2008, 11:02 AMHallsofIvy
Geherally the best way to solve a complicated inequality, like or, equivalently , is to solve the corresponding equation: .

Multiply the entire equation by x(x+1)(x+2): (x+1)(x+2)+ x(x+1)= x(x+2). Multiplying that out, which reduces to . The discriminant for that is [tex]2^2- 4(1)(2)= -4 so there are NO real number roots.

A function can change from ">" to "<" or vice-versa only at points where it is "=" or where it is discontinuous. Since there are no points where it is "=", it can only change at points where it is not continuous: where one of the denominators is 0, that is at x= -1, x= -2, and x= 0.

You can check a single value in each interval having those endpoints to determine whether every point in that interval satisfies the inequality or not.

For example, -3 is in the interval of all numbers less than -2 and [tex]\frac{1}{-3}+ \frac{1}{-3+ 2}= -\frac{1}{3}-1= -\frac{4}{3} math which < is than [tex]\frac{1}{-3+1}= -\frac{1}{2}[tex] so the inequality is satisified for all x< -2. Try, say, x= -3/2, which is between -2 and -1 to check if that interval works, x= -1/2 for the interval (-1, 0) and x= 1 for the interval of numbers larger than 1.